set.seed(1) # ensures reproducibility
B <- 1000 # number of simulations
n <- 500 # sample size per group
p1 <- 0.5 # true proportion in group 1
p2 <- 0.4 # true proportion in group 2
# Create empty vectors to store simulated means and SDs
mean_x1 <- mean_x2 <- numeric(B)
sd_x1 <- sd_x2 <- numeric(B)
# Loop to simulate samples for both groups
for (i in 1:B) {
# Generate random binary outcomes for group 1 (success = 1, failure = 0)
x1 <- rbinom(n, size = 1, prob = p1)
mean_x1[i] <- mean(x1) # sample proportion for group 1
sd_x1[i] <- sd(x1) # sample SD for group 1
# Repeat for group 2
x2 <- rbinom(n, size = 1, prob = p2)
mean_x2[i] <- mean(x2)
sd_x2[i] <- sd(x2)
}Unit III
ECON 3406
Dr. Josh Martin
Differences of Two Proportions
- Like with \(\hat{p}\), the difference of two sample proportions \(\hat{p}_1 - \hat{p}_2\) can be modeled using a normal distribution (when conditions are met).
\(SE = \sqrt{\dfrac{p_1 (1-p_1)}{n_1} + \dfrac{p_2 (1-p_2)}{n_2}}\)
- This standard error comes from the fact that variances of independent variables add, even when subtracting.
Differences of Two Proportions: Standard Errors
When we talk about the spread of an estimate, we’re really talking about variance (the square of the standard error).
If two random variables A and B are independent, then:
\(\text{Var}(A - B) = \text{Var}(A) + \text{Var}(B)\)
- This might seem counterintuitive — but remember:
- Even if you’re subtracting two noisy measurements, the uncertainty (noise) from both still adds up.
- Think of it like using two shaky rulers. Subtracting doesn’t cancel the shakiness — it just combines it!
Differences of Two Proportions: Simulation Setup
We’ll use simulation to understand the sampling distribution of sample proportions for two independent groups.
- Group 1 has a true proportion \(p_1 = 0.5\)
- Group 2 has a true proportion \(p_2 = 0.4\)
- Each group has \(n = 500\) individuals per sample
- We’ll repeat this sampling 1,000 times to observe variation in sample means
Differences of Two Proportions: Simulation
Comparing Theoretical and Empirical Standard Errors
- We can compare:
- the theoretical standard error (from the formula)
- the empirical standard error (from our simulations)
Comparing Theoretical and Empirical Standard Errors
[1] 0.02236068[1] 0.0223613- The first line gives the theoretical SE: \(\sqrt{p(1-p)/n}\)
- The second line gives the empirical SE, based on simulated SDs
- These values should be nearly identical, validating the normal approximation for large \(n\)
Sampling Distribution for One Group (p1 = 0.5)
- We can visualize the distribution of sample proportions across simulations, and overlay a 95% confidence interval around the true mean.
Sampling Distribution for One Group (p1 = 0.5)
Sampling Distribution for One Group (p1 = 0.5)
- Roughly 5% of simulated sample proportions should fall outside this interval — confirming the 95% confidence level’s interpretation.
Sampling Distribution for One Group (p2 = 0.4)
Sampling Distribution for One Group (p2 = 0.4)
- Again, around 5% of simulated estimates will fall outside the interval.
- The spread is slightly narrower than for Group 1 because the variance is smaller.
Combining Two Proportions
- Now that we understand the sampling variation of each group separately, we can combine them just as we would when estimating a difference in proportions:
\(SE_{\hat{p}_1 - \hat{p}_2} = \sqrt{SE_{\hat{p}_1}^2 + SE_{\hat{p}_2}^2}\)
- This formula reflects that variances add, even though we’re subtracting proportions.
Simulated Differences Between Groups
Simulated Differences Between Groups
Simulated Differences Between Groups
Differences of two proportions: Example 1
- Consider an experiment involving patients who underwent cardiopulmonary resuscitation (CPR) following a heart attack and were subsequently admitted to a hospital.
- Patients were randomly assigned to either a treatment group (received a blood thinner) or a control group (no blood thinner).
- The outcome of interest was survival for at least 24 hours.
Differences of two proportions: Example 1
| Survived | Died | Total | |
|---|---|---|---|
| Control | 11 | 39 | 50 |
| Treatment | 14 | 26 | 40 |
| Total | 25 | 65 | 90 |
Differences of two proportions: Example 1
- Create and interpret a 90% confidence interval of the difference for the survival rates in the CPR study.
- \(p_t - p_c = 0.35 - 0.2 = 0.13\)
- \(SE = \sqrt{\dfrac{0.35 (1-0.35)}{40} + \dfrac{0.22 (1-0.22)}{50}} \approx 0.095\)
- \(0.13 \pm 1.645 \times 0.095 = (-0.027, 0.287)\)
Differences of two proportions: Computing in R
pt <- 14/40
pc <- 11/50
nt <- 40
nc <- 50
point_est <- pt - pc
se <- sqrt((pt * (1 - pt) / nt) + (pc * (1 - pc) / nc))
z <- data.frame(
sig_level = c(0.01, 0.05, 0.1),
z_score = c(2.45, 1.96, 1.645)
)
z$min <- point_est - z$z_score * se
z$max <- point_est + z$z_score * se
z sig_level z_score min max
1 0.01 2.450 -0.10396538 0.3639654
2 0.05 1.960 -0.05717230 0.3171723
3 0.10 1.645 -0.02709104 0.2870910Differences of two proportions: Visualizing Confidence Intervals
Differences of two proportions: Interpretation
We are 90% confident that blood thinners change the 24-hour survival rate by between -3 and 29 percentage points for patients similar to those in the study.
Because 0% is within this range, the evidence is inconclusive — we cannot determine whether blood thinners help or harm heart attack patients who have undergone CPR.
Differences of Two Proportions: Example 2
- A 5-year clinical trial evaluated whether fish oil supplements reduce the risk of heart attacks.
- Each participant was randomly assigned to one of two groups:
- Fish Oil group
- Placebo group
- We’ll examine heart attack outcomes across both groups.
Differences of Two Proportions: Example 2
| Group | Heart Attack | No Event | Total |
|---|---|---|---|
| Fish Oil | 145 | 12,788 | 12,933 |
| Placebo | 200 | 12,738 | 12,938 |
Differences of Two Proportions: Example 2
- Construct a 95% confidence interval for the effect of fish oil on heart attack incidence among patients represented by this study.
- Interpret the interval in context:
- What does the direction and width of the interval suggest?
- Is there evidence that fish oil has a meaningful effect on heart attack risk?
Differences of two proportions: Computing in R
Differences of two proportions: Visualizing Confidence Intervals
Differences of two proportions: Interpretation
The point estimate for the effect of fish oil is approximately –0.0043,
meaning heart attacks occurred 0.43 percentage points less often in the fish-oil group than in the placebo group.We are 90% confident that fish oil changes the heart-attack rate by between –0.66 and –0.19 percentage points for patients similar to those in the study.
Because this interval does not include 0, the reduction in heart-attack risk is statistically significant at the 10% (and even 5% an 1%) level.
Practical vs. Statistical Significance
- While statistically significant, the effect size is extremely small — roughly 0.4 fewer heart attacks per 100 individuals.
- In a large clinical sample, even minor effects can reach significance if variability is low.
- From a practical standpoint, such a small reduction may not justify the cost, side effects, or adherence burden of treatment.
More on Two-Proportion Hypothesis Tests
When conducting a two-proportion hypothesis test, the null hypothesis is typically: \(H_0: p_1 - p_2 = 0\)
However, there are cases where we may want to test for a specific difference other than zero.
- For example, suppose we want to test whether: \(H_0: p_1 - p_2 = 0.10\)
In contexts like these, we use the sample proportions \(\hat{p}_1\) and \(\hat{p}_2\) to check the success–failure condition and to construct the standard error.
Differences of Two Proportions: Example 3
A drone quadcopter company is considering a new manufacturer for rotor blades.
The new manufacturer is more expensive but claims that their higher-quality blades are 3% more reliable, meaning that 3% more blades pass inspection compared to the current supplier.
Set up the appropriate hypotheses for this test:
- \(H_0: p_{\text{highQ}} - p_{\text{standard}} = 0.03\)
- \(H_A: p_{\text{highQ}} - p_{\text{standard}} \neq 0.03\)
Differences of Two Proportions: Example 3 (Data)
- A quality control engineer collects samples of 1,000 blades from each manufacturer:
- Current supplier: 899 blades pass inspection
- Prospective supplier: 958 blades pass inspection
- Current supplier: 899 blades pass inspection
- Using these data, evaluate the hypotheses above at a significance level of 5%.
Compute the Point Estimate and Standard Error
Compute Confidence Intervals for Various Significance Levels
z <- data.frame(
sig_level = c(0.01, 0.05, 0.10),
z_score = c(2.45, 1.96, 1.645)
)
z$min <- (point_est - z$z_score * se) - 0.03
z$max <- (point_est + z$z_score * se) - 0.03
z sig_level z_score min max
1 0.01 2.450 0.0009547225 0.05704528
2 0.05 1.960 0.0065637780 0.05143622
3 0.10 1.645 0.0101695994 0.04783040Visualizing Confidence Intervals
Compute and Visualize the z-Statistic
Visualizing the Sampling Distribution
Example 3: Conclusion
- From the standard normal distribution:
- The right-tail area is approximately 0.004
- Doubling for a two-tailed test gives p = 0.008
- Since p = 0.008 < 0.05, we reject the null hypothesis
- We find statistically significant evidence that the higher-quality blades have a pass rate greater than 3% higher than the standard blades — exceeding the company’s claims.
Chi-Squared Distributions: Introduction
\(\chi\) = the greek letter for “chi” (pronounced like “kai”)
The \(\chi^2\) distribution is a continous probability distribution that is widely used in statistical inference.
- Closely related to the standard normal distribution
If a variable \(Z\) has the standard normal distribution, then \(Z^2\) has the \(\chi^2\) distribution with one degree of freedom
Chi-Squared Distributions: Histograms
Chi-Squared Distributions: Definition
- If \(Z_1, Z_2,..., Z_k\) are independent standard normal variables, then…
- \(Z_1^2 + Z_2^2 +...+ Z_k^2\)
- …has a \(\chi^2\) distribution with \(k\) degrees of freedom.
Degrees of Freedom: Concept
A degree of freedom (df) represents the number of independent pieces of information available to estimate something.
Whenever we calculate a statistic, we “use up” some information.
- For example, once we estimate the sample mean, one data point can be perfectly predicted from the others.
- So for a sample of size \(n\), only \((n - 1)\) observations are free to vary when computing the sample variance.
Degrees of Freedom: Intuition
Average (mean): \(\bar{x} = \frac{1}{n} \sum_i^n x_i = \frac{x_1 + ... + x_n}{n}\)
if \(i = 4\), \(x_1 = 8\), \(x_2 = 10\), \(x_3 = 12\) and \(\bar{x} = 10\)…
- … then \(x_4 = 10\)
So even though we had four data points, only three were free to vary — the fourth is determined by the mean.
That’s why when calculating the sample variance, we divide by \(n\) instead of \(n-1\): one degree of freedom has been “used up” in estimating the mean.
Degrees of Freedom: Motivation
- Why it matters:
- Degrees of freedom tell us how much independent information our test or estimate is based on.
- They affect the shape of sampling distributions (like \(t\), \(F\), and \(\chi^2\)), which in turn changes the critical values and p-values we use.
- More degrees of freedom → more information → the distribution becomes narrower and more normal-looking.
- Big idea:
- Degrees of freedom link sample size, uncertainty, and the reliability of inference — they remind us that every time we estimate something, we “spend” information.
Chi-Squared Distributions: Properties
- mean: \(\mu = k\)
- variance: \(\sigma^2 = 2k\)
- mode occurs at \(\mu - 2\)
Chi-Squared Distributions: Multiple Degrees of Freedom
Chi-Squared Distributions: Multiple Degrees of Freedom
Chi-Squared Distributions: Multiple Degrees of Freedom
Chi-Squared Distributions: Multiple Degrees of Freedom
Chi-Squared Tests
- In this section, we develop a method for assessing a null model when data are binned into categories.
- This technique is commonly used in two settings:
- To determine whether a sample is representative of a larger population across several groups.
- To evaluate whether observed data follow a specified theoretical distribution (e.g., normal, geometric, Poisson).
- Both scenarios rely on the same statistical tool — the chi-squared test — which allows us to examine all bins simultaneously, rather than testing one or two categories in isolation.
Chi-Squared Tests: Jurors Example
- Consider data from a random sample of 275 jurors in a small county.
- Each juror identified their racial group.
- We would like to test whether this sample is racially representative of the eligible population of jurors (i.e., registered voters).
Chi-Squared Tests: Jurors Example
Chi-Squared Tests: Jurors Example
The sample proportions of jurors by race do not perfectly match those in the voter population.
The question is whether these differences are large enough to suggest that the jury was not randomly selected.
We can summarize this as a formal hypothesis test:
- H₀ (Null Hypothesis): Jurors are randomly selected; any differences reflect natural sampling variation.
- Hₐ (Alternative Hypothesis): Jurors are not randomly selected; the differences are too large to attribute to chance alone.
- H₀ (Null Hypothesis): Jurors are randomly selected; any differences reflect natural sampling variation.
Chi-Squared Tests: Building the Test Statistic
- In previous hypothesis tests, our test statistic took the general form:
\[ \frac{\text{Point Estimate} - \text{Null Value}}{\text{Standard Error of Point Estimate}} \]
- This approach involved two key steps:
- Measuring the difference between the observed value and what we would expect if the null hypothesis were true.
- Standardizing that difference using the standard error of the point estimate.
- Measuring the difference between the observed value and what we would expect if the null hypothesis were true.
Chi-Squared Test Statistic (Preview)
For categorical data, the test statistic combines the squared, standardized deviations across all categories:
\[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} \]
Where:
- \(O_i\) = observed count in category \(i\)
- \(E_i\) = expected count under the null hypothesis
- \(O_i\) = observed count in category \(i\)
This statistic follows a chi-squared distribution with \((k - 1)\) degrees of freedom, where \(k\) is the number of categories.
Chi-Squred Tests: Calculating X^2
Chi-Squred Tests: Simulations
set.seed(1)
B <- 200000
Y <- t(rmultinom(B, size = n, prob = p))
m <- matrix(E, B, k, byrow=TRUE)
colnames(m) <- colnames(Y) <- x$race
head(Y) white black hispanic other
[1,] 196 20 29 30
[2,] 205 23 25 22
[3,] 203 15 36 21
[4,] 199 25 32 19
[5,] 202 16 34 23
[6,] 203 22 30 20 white black hispanic other
[1,] 198 19.25 33 24.75
[2,] 198 19.25 33 24.75
[3,] 198 19.25 33 24.75
[4,] 198 19.25 33 24.75
[5,] 198 19.25 33 24.75
[6,] 198 19.25 33 24.75Chi-Squred Tests: Simulations
- The p-value of 0.117 indicates that, if jurors were selected randomly according to the county’s racial composition, there is about an 11.7% chance of observing differences in juror counts at least as extreme as those in the sample
Chi-Squred Tests: Visualizing the p-value
Chi-Squred Tests: Attempting a More Familiar Approach (Incorrect)
Chi-Squred Tests: Attempting a More Familiar Approach (Incorrect)
Why the “More Familiar Way” Is Incorrect
- This approach standardizes each category as if they were independent samples.
- When one category increases, another must decrease, introducing negative covariance.
- Ignoring this dependence effectively uses too many degrees of freedom, overstating variability.
- When one category increases, another must decrease, introducing negative covariance.
- The correct chi-squared test adjusts for this by using \((k - 1)\) degrees of freedom.
Chi-Squared Tests: Stock Market Example
- The chi-squared framework can also be used to evaluate how well a statistical model fits observed data.
- Suppose we analyze daily stock returns from the S&P 500 over a 10-year period to test whether market movements are independent across days.
- In other words, does knowing whether the market went up or down yesterday help predict what happens today?
- In other words, does knowing whether the market went up or down yesterday help predict what happens today?
- This question has clear implications for traders:
- If past information helps forecast future returns, that knowledge could provide a trading advantage.
Chi-Squared Tests: Stock Market Example
To explore this, we can label each trading day as “Up” or “Down”, and record the number of days between Up days to study the waiting-time pattern.
If daily market movements are truly independent, the waiting time between positive (Up) days should follow a geometric distribution.
The geometric distribution gives the probability of waiting \(k\) trials to observe the first success: \(P(X = k) = (1 - p)^{k - 1}p\)
- \(p\) = probability that a given day is an Up day
- \(X\) = number of days until the next Up day
- \(p\) = probability that a given day is an Up day
Chi-Squred Tests: Computing Expected Counts
Chi-Squred Tests: Comparing Observed and Expected Rallies
Chi-Squred Tests: Simulating the Chi-Squared Distribution
Chi-Squred Tests: Visualizing the p-Value
Intro to t-distributions
We rarely know the true population mean (\(\mu\)) — we estimate it from a sample.
When the population standard deviation (\(\sigma\)) is unknown (which is most of the time), we rely on the \(t\)-distribution instead of the normal model.
The \(t\)-test helps us quantify uncertainty and make inferences about population parameters using sample data.
t-distributions
t-distributions: Degrees of Freedom
Intuition of t-distributions
Using the t-Distribution
- We use a t-distribution with (df = n - 1) to model the sample mean when the sample size is \(n\).
- As the number of observations increases, the degrees of freedom (df) also increase.
- With larger df, the t-distribution more closely resembles the standard normal distribution.
- When \(df \approx 30\) or greater, the t-distribution is nearly indistinguishable from the normal model.
- In general, the degrees of freedom describe the shape of the t-distribution — larger df means the distribution becomes narrower and more normal-like.
Example: Mercury in Dolphins
- Let’s apply the t-distribution in a real-world example involving mercury content in dolphin muscle.
- Elevated mercury concentrations are a serious concern for both dolphins and humans who consume them.
- We’ll construct a confidence interval for the average mercury concentration using a sample of 19 Risso’s dolphins from the Taiji area in Japan.
- Elevated mercury concentrations are a serious concern for both dolphins and humans who consume them.
Example: Mercury in Dolphins
| Statistic | Value |
|---|---|
| n | 19 |
| 𝑥̄ (mean) | 4.4 |
| s (SD) | 2.3 |
| Min | 1.7 |
| Max | 9.2 |
\[\text{point estimate} \pm t^*_{df} \times \frac{s}{\sqrt{n}}\]
t-statistics and p-values
t-statistics and confidence intervals
Example 2: Are Runners Getting Faster?
Is the typical U.S. runner getting faster or slower over time?
We’ll explore this question using data from the Cherry Blossom 10-Mile Run, held each spring in Washington, D.C.
- In 2006, the average finish time for all runners was 93.29 minutes (about 93 minutes and 17 seconds).
- We’ll use a sample of 100 participants from the 2017 race to test whether the average finish time has changed.
- In other words, are runners getting faster or slower, or has there been no meaningful change?
Example 2: Calculating the t-statistic
\[t = \dfrac{\text{estimate} - \text{null}}{sd / \sqrt{n}}\]
Example 2: Vizualizing the p-value
Example 2: Interpreting the Results
- Because the p-value is smaller than 0.05, we reject the null hypothesis.
- The data provide strong evidence that the average run time in 2017 differs from the 2006 average.
- Since the observed mean time is higher than the 2006 mean, and we’ve rejected the null, we conclude that: Runners were slower on average in 2017 than in 2006.
- The data provide strong evidence that the average run time in 2017 differs from the 2006 average.
Difference of Two Means
- In this section, we consider the difference between two population means \(\mu_1 - \mu_2\) when the data are not paired.
- As with a single sample, we must check certain conditions to ensure we can use the t-distribution.
- The point estimate for the difference is \(\bar{x}_1 - \bar{x}_2\).
- A new standard error formula is used to reflect variation across both samples.
- Aside from these two differences, the procedures are nearly identical to the one-mean case.
- As with a single sample, we must check certain conditions to ensure we can use the t-distribution.
Example 3: ESC and Heart Recovery
- Does treatment using embryonic stem cells (ESCs) help improve heart function after a heart attack?
- An experiment tested this question in sheep that had experienced a heart attack.
- Each sheep was randomly assigned to either an ESC treatment group or a control group.
- Researchers measured the change in heart pumping capacity after treatment.
- A positive value indicates stronger recovery.
- Our goal is to construct a 95% confidence interval for the effect of ESC treatment on heart recovery compared to the control group.
- An experiment tested this question in sheep that had experienced a heart attack.
Example 3: Distribution
Example 3: Summary Statistics
trmt before after diff
1 ctrl 35.25 29.50 -5.75
2 ctrl 36.50 29.50 -7.00
3 ctrl 39.75 36.25 -3.50
4 ctrl 39.75 38.00 -1.75
5 ctrl 41.75 37.50 -4.25
6 ctrl 45.00 42.75 -2.25x <- aggregate(diff ~ trmt, y,
function(x) c(mean = mean(x),
sd = sd(x),
n = length(x)))
x <- as.data.frame(do.call(cbind, x))
x[,2:4] <- apply(x[,2:4], 2, as.numeric)
x trmt mean sd n
1 ctrl -4.333333 2.764168 9
2 esc 3.500000 5.172040 9Example 3: Standard Errors
\(SE = \sqrt{SE_1 + SE_2} = \sqrt{\frac{SD_1^2}{n_1} + \frac{SD_2^2}{n_2}}\)
Example 4: Smoking and Birth Weight
The ncbirths dataset contains a random sample of 150 mothers and their newborns in North Carolina collected over the course of a year.
- We focus on two key variables:
- weight — the newborn’s birth weight (in pounds or ounces)
- smoke — whether the mother smoked during pregnancy
- weight — the newborn’s birth weight (in pounds or ounces)
- Our question: Is there convincing evidence that newborns of mothers who smoke have a different average birth weight than those of non-smoking mothers?
Example 4: Randomization
Example 4: Summary Statistics
Example 4: t-test and p-value
ANOVA: Comparing More Than Two Means
Sometimes we want to compare means across multiple groups.
We might be tempted to make pairwise comparisons — for example, comparing
- Group 1 vs. Group 2
- Group 1 vs. Group 3
- Group 2 vs. Group 3
- Group 1 vs. Group 2
ANOVA: Comparing More Than Two Means
- However, this approach can be misleading. With many groups, making multiple comparisons increases the chance of finding a difference by random chance, even if no real differences exist.
- Instead, we use a holistic test — the Analysis of Variance (ANOVA) — to assess whether there is evidence that at least one group mean differs from the others.
- In short: ANOVA saves the day by allowing us to test for overall mean differences across several groups at once.
Core Idea of ANOVA
We will learn a new method called analysis of variance (ANOVA) and a new test statistic called \(F\)
ANOVA uses a single hypothesis test to check whether the means across many groups are equal
- \(H_0 = \mu_1 = \mu_2 = ... = \mu_k\)
- \(H_A =\) at least one mean is different
ANOVA Example: Exam Scores Across Classes
- College departments often offer multiple sections of the same introductory course each semester to meet high student demand.
- Suppose a statistics department runs three lectures (A, B, and C) of an introductory statistics course.
- We want to know whether there are statistically significant differences in the average first exam scores across these three classes.
- ANOVA allows us to test whether any of the class means differ, rather than relying on multiple pairwise comparisons.
ANOVA: SSG
Statisticians measure variation using squared differences. The sum of squares for the groups (SSG) measures the differences between the group means \(\bar{x}_i\) and the grand mean \(\bar{x}\) like so:
\[SSG = \sum_{i=1}^k n_i (\bar{x}_i - \bar{x})^2\]
ANOVA: SSG
ANOVA: MSG
The mean square between groups (MSG) is calculated using both the groups’ degrees of freedom \(df_G\) and the sum of squares between groups SSG like so:
\[MSG = \dfrac{1}{df_G} SSG\]
ANOVA: SSE
The sum of squared errors (SSE) is the sum of the squares of the differences between each sample’s observations with each respective sample mean
The calculation is simplified by first finding the variance \(s_i^2\) of each sample:
ANOVA: SSE
\[SSE = \sum_{i = 1}^k (n_i - 1) s_i^2\]
ANOVA: MSE
\[MSE = \dfrac{1}{df_E} SSE\]
ANOVA: F-Statistic
\[F = \dfrac{MSG}{MSE}\]
- We now have comparable measures of variability both between the groups (MSG) and within the groups (MSE).
- If variation among group means is simply random, F is typically close to 1
ANOVA: p-value from the F-statistic
ANOVA: Vizualizing the F-statistic
ANOVA: p-value from the F-statistic (easy)
score group
1 90 a
2 87 a
3 88 a
4 78 a
5 87 b
6 85 b
7 80 b
8 95 c
9 93 c
10 90 c
11 88 c
12 85 c Df Sum Sq Mean Sq F value Pr(>F)
group 2 84.12 42.06 2.181 0.169
Residuals 9 173.55 19.28 ANOVA: Statistical Signifiance
semester sex exam1 exam2 exam3 course_grade
1 2000-1 Man 84.5 69.5 86.5 76.2564
2 2000-1 Man 80.0 74.0 67.0 75.3882
3 2000-1 Man 56.0 70.0 71.5 67.0564
4 2000-1 Man 64.0 61.0 67.5 63.4538
5 2000-1 Man 90.5 72.5 75.0 72.3949
6 2000-1 Man 74.0 78.5 84.5 71.4128 Df Sum Sq Mean Sq F value Pr(>F)
semester 5 1261 252.14 2.719 0.0208 *
Residuals 227 21053 92.74
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1